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frosja888 [35]
3 years ago
15

If it’s right i’ll give brainliest :)

Mathematics
2 answers:
SpyIntel [72]3 years ago
7 0

Answer:

x = 126

Step-by-step explanation:

-\frac{2}{7}x - \frac{8}{21}x + \frac{1}{3}x = -42

-\frac{2}{7}x (multiply numerator and denominator by 3)

-\frac{6}{21}x

\frac{1}{3}x (multiply numerator and denominator by 7)

\frac{7}{21}x

-\frac{6}{21}x - \frac{8}{21}x + \frac{7}{21}x = -42

-\frac{7}{21}x = -42 (multiply everything by -21)

7x = 882 (divide everything by 7)

x = 126

fomenos3 years ago
6 0

Answer:

x=126

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
45 divided by 2531 HELP IDK HOW TO DO THIS.
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Consider the system of equations. 5x – 2y = 6 –10x + 4y = –12 Check all the statements that are true. a)the lines coincide. b)Th
QveST [7]

Answer:

a) the lines coincide, c) the slopes are equal, d) the y-intercept are equal, e) the system has an infinite number of solutions.

Step-by-step explanation:

Each equation is a multiple of the other one. Thus, the system of equation has an infinite of solutions, reports same slopes and y-intercepts and coinciding lines.

5 0
4 years ago
At a college campus, random drug tests are carried out to prevent use of prohibited drugs by athletes. Of all the athletes on ca
saw5 [17]

Answer:

a) 0.3064 = 30.64% probability that athlete will test positive for prohibited drug use.

b) 0.2693 = 26.93% probability that this drug user never used the prohibited drug.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

A. If an athlete is selected at random, what is the probability that athlete will test positive for prohibited drug use?

100 - 5 = 95% of 8%

100 - 13 = 87% of 17%

11% of 75%

So

p = 0.95*0.08 + 0.87*0.17 + 0.11*0.75 = 0.3064

0.3064 = 30.64% probability that athlete will test positive for prohibited drug use.

B. If an athlete tests positive after a drug test, what is the probability that this drug user never used the prohibited drug?

Conditional Probability

Event A: Tests Positive

Event B: Never used the drug.

0.3064 = 30.64% probability that athlete will test positive for prohibited drug use

This means that P(A) = 0.3064

Probability of testing positive while never using the drug.

11% of 75%. So

P(A \cap B) = 0.11*0.75 = 0.0825

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0825}{0.3064} = 0.2693

0.2693 = 26.93% probability that this drug user never used the prohibited drug.

8 0
3 years ago
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